A bullet of mass 0.00112 kg and speed v passes completely through a pendulum bob of mass 84.6 kg. The bullet emerges with a speed v/2. The pendulum bob is suspended by a stiff rod of length 1.08 m and negligible mass.
The acceleration of gravity is 9.8 m/s62
What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?
Answer in units of m/s.

To find the minimum value of \( v \) such that the pendulum bob barely swings through a complete vertical circle, we need to apply the principles of conservation of momentum and mechanical energy.

### Step 1: Conservation of Linear Momentum
Since the bullet passes through the pendulum bob, there is an inelastic interaction. The system conserves linear momentum during the bullet’s interaction with the bob. The conservation of momentum is written as:

\[
m_b v = m_b \frac{v}{2} + M V
\]

Where:
- \( m_b = 0.00112 \, \text{kg} \) is the mass of the bullet.
- \( M = 84.6 \, \text{kg} \) is the mass of the pendulum bob.
- \( v \) is the initial velocity of the bullet.
- \( V \) is the velocity of the pendulum bob just after the bullet passes through it.

Simplifying this equation:

\[
0.00112 v = 0.00112 \frac{v}{2} + 84.6 V
\]

\[
0.00112 v - 0.00056 v = 84.6 V
\]

\[
0.00056 v = 84.6 V
\]

\[
V = \frac{0.00056 v}{84.6}
\]

\[
V = \frac{v}{151071.43}
\]

### Step 2: Energy Required for Complete Vertical Circle
For the pendulum bob to complete a vertical circle, the velocity \( V \) at the lowest point must be enough to provide the required centripetal force at the top of the circle.

The condition for completing the vertical circle is that the kinetic energy at the lowest point must be enough to raise the bob to a height of \( 2L \), where \( L = 1.08 \, \text{m} \) is the length of the pendulum rod. This means the total mechanical energy at the bottom equals the potential energy at the top of the swing. We use conservation of mechanical energy:

\[
\frac{1}{2} M V^2 = M g (2L)
\]

Where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Simplifying:

\[
\frac{1}{2} V^2 = g(2L)
\]

\[
V^2 = 4gL
\]

\[
V = \sqrt{4gL}
\]

Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( L = 1.08 \, \text{m} \):

\[
V = \sqrt{4 \times 9.8 \times 1.08}
\]

\[
V = \sqrt{42.336} \approx 6.506 \, \text{m/s}
\]

### Step 3: Solve for \( v \)
Now we know that the minimum velocity \( V \) of the pendulum bob is \( 6.506 \, \text{m/s} \). Using the earlier expression \( V = \frac{v}{151071.43} \), we can solve for \( v \):

\[
6.506 = \frac{v}{151071.43}
\]

\[
v = 6.506 \times 151071.43
\]

\[
v \approx 982.86 \, \text{m/s}
\]

### Final Answer:
The minimum value of \( v \) such that the pendulum bob will barely swing through a complete vertical circle is approximately:

\[
v \approx 982.86 \, \text{m/s}
\]

---

Would you like more details on any step, or do you have any other questions?

Here are 5 related questions for deeper understanding:
1. How does the mass of the pendulum bob affect the minimum velocity of the bullet?
2. What is the relationship between the height of the swing and the velocity of the bob?
3. How would the result change if the bullet only grazed the pendulum bob instead of passing through?
4. Can the pendulum complete a vertical circle if the rod has mass? Why or why not?
5. How does the length of the pendulum affect the minimum required velocity for the bullet?

**Tip**: In problems involving energy and momentum, always carefully determine which principles (momentum or energy) to apply in different parts of the system.

A bullet of mass 0.00112 kg and speed v passes completely through a pendulum bob of mass 84.6 kg. The bullet emerges with a speed v/2. The pendulum bob is suspended by a stiff rod of length 1.08 m and negligible mass. The acceleration of gravity is 9.8 m/s². What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Answer in units of m/s.

Learn how to calculate the minimum speed of a bullet needed to make a pendulum bob swing through a complete vertical circle. This problem uses principles of conservation of momentum and mechanical energy, with detailed step-by-step explanations. Suitable for college-level physics students.

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